r - 按组计算基于两列的唯一行数

Question

我在r中有一个data.table

    col1 col2 col3   col4
 1:  5.1  3.5  1.4 setosa
 2:  5.1  3.5  1.4 setosa
 3:  4.7  3.2  1.3 setosa
 4:  4.6  3.1  1.5 setosa
 5:  5.0  3.6  1.4 setosa
 6:  5.1  3.5  3.4    eer
 7:  5.1  3.5  3.4    eer
 8:  5.1  3.2  1.3    eer
 9:  5.1  3.5  1.5    eer
10:  5.1  3.5  1.4    eer


DT <- structure(list(col1 = c(5.1, 5.1, 4.7, 4.6, 5, 5.1, 5.1, 5.1, 
5.1, 5.1), col2 = c(3.5, 3.5, 3.2, 3.1, 3.6, 3.5, 3.5, 3.2, 3.5, 
3.5), col3 = c(1.4, 1.4, 1.3, 1.5, 1.4, 3.4, 3.4, 1.3, 1.5, 1.4
), col4 = structure(c(1L, 1L, 1L, 1L, 1L, 4L, 4L, 4L, 4L, 4L), .Label = c("setosa", 
"versicolor", "virginica", "eer"), class = "factor")), .Names = c("col1", 
"col2", "col3", "col4"), row.names = c(NA, -10L), class = c("data.table", 
"data.frame"))

我想为 col1的每个值计算 col2和 col4的唯一(不同)组合。

预期输出为

   col1 col2 col3   col4 count
 1:  5.1  3.5  1.4 setosa     4
 2:  5.1  3.5  1.4 setosa     4
 3:  4.7  3.2  1.3 setosa     4
 4:  4.6  3.1  1.5 setosa     4
 5:  5.0  3.6  1.4 setosa     4
 6:  5.1  3.5  3.4    eer     2
 7:  5.1  3.5  3.4    eer     2
 8:  5.1  3.2  1.3    eer     2
 9:  5.1  3.5  1.5    eer     2
10:  5.1  3.5  1.4    eer     2

如何仅使用1个data.table语法执行此操作？

Answer 1

我必须先进行几次尝试，然后最终完成。有什么好处吗？

DT[, count:=nrow(unique(.SD)), by=col4, .SDcols=c("col1","col2")]
DT
    col1 col2 col3   col4 count
 1:  5.1  3.5  1.4 setosa     4
 2:  5.1  3.5  1.4 setosa     4
 3:  4.7  3.2  1.3 setosa     4
 4:  4.6  3.1  1.5 setosa     4
 5:  5.0  3.6  1.4 setosa     4
 6:  5.1  3.5  3.4    eer     2
 7:  5.1  3.5  3.4    eer     2
 8:  5.1  3.2  1.3    eer     2
 9:  5.1  3.5  1.5    eer     2
10:  5.1  3.5  1.4    eer     2
> 

和相同但更快的感谢Procrastinatus在下面的评论:

DT[, count:=uniqueN(.SD), by=col4, .SDcols=c("col1","col2")]